Chapter 9 Calculations from Chemical Equations (Stoichiometry) Objectives: The Mole-Ratio Method Mole-Mole Calculations Mole-Mass Calculations Mass-Mass Calculations Limiting-Reactant and Percent Yield Calculations

A Short Review Molar Mass Molecules and Moles Balanced Equations Sum of the atomic masses of all atoms in a molecule Molecules and Moles A molecule is the smallest unit of a substance A mole is Avogadro’s number (6.022 x 1023) molecules of that substance Balanced Equations Equations must be balanced Number in front of a formula in a balanced chemical equation represent the number of moles of that substance

The Mole-Ratio Method Stoichiometry Mole ratio quantitative relationships among reactants and products Mole ratio Ratio between the number of moles of any two species involved in a chemical reaction 2H2 + O2  2H2O Six mole ratios can be written: 2 mol H2 1 mol O2 2 mol H2 2 mol H2O 1 mol O2 2 mol H2 1 mol O2 2 mol H2O 2 mol H2O 2 mol H2 2 mol H2O 1 mol O2

The Mole-Ratio Method = 8.0 mol H2O Use the mole ratio to convert number of moles of one substance to number of moles of another substance How many moles of H2O can be obtained from 4.0 moles of O2? 4.0 mol O2 2 mol H2O 1 mol O2 = 8.0 mol H2O

The Mole-Ratio Method Three basic steps: Convert the quantity of starting substance to moles (if it is not given in moles) Convert the moles of starting substance to moles of desired substance Convert the moles of desired substance to the units specified in the problem Grams of B Grams of A Moles of A Moles of B Atoms or Molecules hof A Atoms or Molecules of A

Moles Of Given Moles Of Unknown Quantity of Given (mass, atoms, molecules) Quantity of Unknown (mass, atoms, molecules) Convert to Moles Convert to desired units Moles Of Given Find molar ratio Moles Of Unknown

Mole-Mole Calculations Quantity of given substance in moles Quantity of desired substance requested in moles How many moles of carbon dioxide will be produced by the complete reaction of 2.0 mol of glucose (C6H12O6) according to the following reaction? C6H12O6 + 6O2  6CO2 + 6H2O 2.0 mol C6H12O6 6 mol CO2 1 mol C6H12O6 = 12 mol CO2

Mole-Mole Calculations N2H4 + 2H2O2  N2 + 4H2O If you have 3 moles of N2H4 how many moles of N2 will you produce? 3 mol N2H4 1 mol N2 = 1 mol N2H4 How many molecules of H2O2 are needed to produce 3 moles of H2O? 3 mol H2O 2 mol H2O2 6.022 X 1023 molecules 4 mol H2O 1 mol H2O2 3 mol N2 = 9 x 1023 molecules H2O2

Mole-Mass Calculations Calculate the mass of one substance when given moles of another >> OR << Calculate the moles of one substance when given mass of another Mass of substance 1 to moles of substance 1 to moles of substance 2 (using mole ratio)

Mole-Mass Calculations 2KClO3  2KCl + 3O2 How many grams of potassium chlorate are needed to produce 5 moles of oxygen? How many moles of potassium chloride can be produced from 100.0 g of potassium chlorate? Do these problems on the board! 5 moles O2 | 2 mol KClO3 | 122.55 g KClO3 = 408.5 g KClO3 | 3 mol O2 | 1 mol KClO3 100.0 g KClO3 | 1 mol KClO3 | 2 mol KCl = 0.8160 mol KCl | 122.55 KClO3 | 2 mol KClO3

Mass-Mass Calculations Calculate the mass of one substance when given the mass of another Mass of substance 1  moles of substance 1  moles of substance 2  mass of substance 2

Mass-Mass Calculations 2AgNO3 + H2S  Ag2S + 2HNO3 How many grams of silver nitrate are required to produce 250.0 grams of silver sulfide? How many grams of nitric acid will be produced if 325 grams of H2S is used? 250 g Ag2S | 1 mol Ag2S | 2 mol AgNO3 | 169.9 g AgNO3 = 342.7 g AgNO3 | 247.9 g Ag2S | 1 mol Ag2S | 1 mol AgNO3 325 g H2S | 1 mol H2S | 2 mol HNO3 | 63.02 g HNO3 = 1201.6 g = 1200 g HNO3 | 34.09 g H2S | 1 mol H2S | 1 mol HNO3

Mass-Mass Calculations What mass of water is produced by the complete combustion of 225.0 g of butane (C4H10)? Write reaction Balance Solve problem 2C4H10 + 13O2  8CO2 + 10H2O 225.0 g C4H10 | 1 mol C4H10 | 10 mol H2O | 18.02 g H2O = 348.8 g H2O | 58.12 g C4H10 | 2 mol C4H10 | 1 mol H2O

Part One Homework Paired Exercises # 3-19 odd Additional Exercises #33 & 37 Enjoy your Thanksgiving!!! See you Tuesday, December 2nd Be prepared for a QUIZ over this information! 

Limiting Reactants Often, quantities of reactants are not perfect One may be left over Amount of product formed depends on reactant that is not in excess Limiting reactant One batch of chocolate chip cookies requires 4 eggs, 4 cups of flour, and 12 oz. of chocolate chips and produces 45 cookies. How many batches can be made from 1 dozen eggs, 15 cups of flour, and 36 oz. of chocolate chips? How many cookies will be produced from 8 eggs, 4 cups of flour, and 24 oz. of chocolate chips? 3 batches (eggs are limiting reactant) Flour is limiting reactant, only 1 batch, so 45 cookies

Limiting Reactants H2 + Cl2  2HCl How many grams of hydrogen chloride can be produced from 0.490 g of hydrogen and 50.0 g of chlorine? Calculate mass-mass for BOTH reactants The one with the LEAST product is your limiting reactant (and thus your answer) 0.490 g H2 | 1 mol H2 | 2 mol HCl | 36.46 g HCl = 17.72 g HCl | 2.016 mol H2 | 1 mol H2 | 1 mol HCl 50.0 g Cl2 | 1 mol Cl2 | 2 mol HCl | 36.46 g HCl = 51.42 g HCl | 70.9 g Cl2 | 1 mol Cl2 | 1 mol HCl Chlorine is limiting reactant…answer is 51.42 g HCl

Limiting Reactants If 36.5 g HCl is reacted with 85.6 g of Ba(OH)2 which is the limiting reactant? How much BaCl2 will be produced? Barium hydroxide is the limiting reactant 104 g BaCl2 will be produced

Percent Yield Thus far, quantities represent maximum yield (100%) Actual yield in lab is not often 100% Side reactions Many reactions reversible Poor lab skills Expected (theoretical) yield calculated amount of product that can be obtained Actual yield Amount actually obtained

Percent Yield Percent yield Ratio of actual yield to theoretical yield Actual yield x 100 = percent yield Expected yield Determine the percent yield for the reaction between 2.80 g Al(NO3)3 and excess NaOH if 0.966 g Al(OH)3 is recovered. (your second product is NaNO3 Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3

Percent Yield A student places an iron nail with a mass of 2.32 g into a flask of CuSO4. The nail reacts completely, leaving a quantity of copper metal in the bottom of the flask. The student finds the mass of the recovered copper to be 2.51 g. The equation for this reaction is: Fe + CuSO4  FeSO4 + Cu What is the expected yield? What is the percent yield?

Percent Yield When octane (C8H18) is burned in oxygen, carbon dioxide and water are produced. If 320 g of octane are burned and 392 g of water are recovered, what is the percent yield of the experiment? First, write and balance the equation Then, answer the question

Percent Yield 4Al + 3O2  2Al2O3 If 0.25 mol Al and 0.40 mol O2 are reacted, which is the limiting reactant? If a student reports a percent yield of 75.5%, how much product (in grams) did she recover?

Part Two Homework Questions # 1 & 2 Paired Exercises # 21, 23, 27 & 29 Additional Exercises # 31 & 41 Be prepared for TEST on Thursday Last EXAM of the quarter… Remember, EC due December 9th